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My works:

$x^2$ can be very large if x is large, thus the function has lose-of-significance error and we need to reformulate it. $$ f(x)=\sqrt{4x^2+x}-2x=\sqrt{x(4x+1)}-2\sqrt{x}\sqrt{x}=\sqrt{x}(\sqrt{4x+1}-2\sqrt{x})$$ The latter will not have any lose-of-significance error in the evaluation.

I am not sure if I did this correctly and I do not know how to compute the limit...

Micah
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walterudoing
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3 Answers3

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I would factor out the $2x$ from both terms, and at a crucial point use the Binomial expansion for $(1+a)^{1/2}$: \begin{align} \sqrt{4x^2+x}-2x&=2x\left[\frac{\sqrt{4x^2+x}}{\sqrt{4x^2}}-1\right]\\ &=2x\left[\sqrt{1+\frac1{4x}}-1\right]\\ &=2x\left[-1+1+\frac12\cdot\frac1{4x}-\frac18\left(\frac{1}{4x}\right)^2+\cdots\right]\\ &=2x\left[ \frac1{8x}-\frac1{128x^2}+\cdots \right]\\ &=1/4-\text{(small)} \end{align}

Lubin
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  • Thanks. Can you please tell me whether the accurate value of this function is correct or not? – walterudoing Dec 08 '13 at 04:35
  • What is the meaning of the expression “accurate value”? Or, what form of description of this accurate value are you looking for? – Lubin Dec 08 '13 at 16:24
  • Formula that does not involve lose-of-significance error in the evaluation. There is a problem that is similar to this one:Question 8 – walterudoing Dec 08 '13 at 17:27
  • In view of the form of the acceptable answer to Question 8, I think that saying the limit is $1/4$ would suffice. – Lubin Dec 08 '13 at 17:38
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Setting $\frac1x=h,$

$$\lim_{x\to\infty^+}\sqrt{4x^2+x}-2x=\lim_{h\to0}\frac{\sqrt{4+h}-2}h\ \ \ \ (1)$$

Method $1:$

Now, $$\frac{\sqrt{4+h}-2}h=\frac{(4+h)-2^2}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}\text{ if }h\ne0$$

Here as $h\to0,h\ne0$

Can you take it from here?

Method $2:$

This has strong resemblance with your method

$\displaystyle \sqrt{4+h}=2\left(1+\frac h4\right)^{\frac12}=2\left(1+\frac12\cdot\frac h4+O(h^2)\right)$ (Using Maclaurin series )

$\displaystyle\implies \sqrt{4+h}-2=\frac h4+O(h^2) $

Now, I leave the rest for you to complete

Method $3:$

$$\lim_{h\to0}\frac{\sqrt{4+h}-2}h=\lim_{h\to0}\frac{\sqrt{4+h}-\sqrt4}h=\frac{d(\sqrt x)}{dx}_{(\text{ at } x=4)}=\frac1{2\sqrt x}_{(\text{ at } x=4)}$$

Method $4:$

Apply L'Hôpital's rule on $(1)$

lab bhattacharjee
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Note that $$\sqrt{4x^2+x}-2x = \dfrac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x} \cdot \sqrt{4x^2+x}-2x = \dfrac{4x^2+x-4x^2}{\sqrt{4x^2+x}+2x} = \dfrac{x}{\sqrt{4x^2+x}+2x}$$ Hence, use $\dfrac{x}{\sqrt{4x^2+x}+2x}$ for computation purposes, especially when $x$ is large. There is no catastrophic cancellation of digits as opposed to evaluating $\sqrt{4x^2+x}-2x$ directly.

  • That can be estimated even closer - $\sqrt{4x^2+x}\approx 2x+\frac{1}{4}$. – Thomas Andrews Dec 08 '13 at 04:20
  • I think that this post addresses the problem of loss of significance error when "x" takes very large errors. If using a computer, where using a bunch a terms from series expansions could not be effcetive, I think that the best solution is what user17762 proposes. – Claude Leibovici Dec 08 '13 at 04:59
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    Downvoters why don't you pinpoint the mistake when its not obvious – lab bhattacharjee Dec 08 '13 at 05:00